Answer:
Explanation:
From the question we are told that:
Slope
initial Concentration
Time
Generally the equation for Raw law is mathematically given by
The partial pressure of argon in the atmosphere is 0.00934 atm. Calculate the partial pressure in mmHg andtorr . Round each of y
1 answer:
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<h3>Answer:</h3>
495 g K₃N
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
3.77 mol K₃N
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
495.039 g K₃N ≈ 495 g K₃N
Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include
, four electron domains, as in ethane
, three electron domains, as in ethene
, two electron domains, as in ethyne
Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.
- Carbon atoms with a
- Carbon atoms with a
- Carbon atoms with a
Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.
Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible